The differences in cross sections are due to differences in wave functions and energy states:
- Slow neutron-proton elastic (and hence total) scattering is enhanced by the existence of a virtual $np$ state.
- Slow neutron capture by a deuteron ($^2\mathrm{H}$) is strongly suppressed because symmetry requirements prevent the initial S-wave $nd$ state from coupling to the dominant S and D components of the triton ($^3\mathrm{H}$ nucleus) wave function.
Understanding even two- and three-nucleon low energy interactions is challenging and has driven much theoretical work that can fit the observed data, but I don't believe there is any trivial explanation why the virtual $np$ state and the triton state fractions have the values they do. (For some examples of recent theoretical work on two- and three-nucleon interactions, see the publications of the Low Energy Nuclear Physics International Collaboration.)
Total Cross-Sections
Because of the lack of low energy $np$ and $nd$ states, their total $np$ and $nd$ cross-sections below $\sim 10$ MeV are completely dominated by elastic scattering.
If there were no bound or unbound nuclear states, the cross section for neutron elastic scattering from nuclei could be naively estimated by the size of the nuclei and the range of the strong nuclear force, i.e.
$$
\sigma_{np}\,\mathrm{(naive)}\sim 4 \pi \left(r_n+r_p +\frac{\hbar c}{m_\pi}\right)^2 \sim 1.2\ \mathrm{b}\\
\sigma_{nd}\,\mathrm{(naive)}\sim 4 \pi \left(r_n+r_d +\frac{\hbar c}{m_\pi}\right)^2 \sim 2.3\ \mathrm{b}
$$
where the radii of the proton, neutron, and deuteron are $r_p\approx 0.84$, $r_n\approx 0.8$ , and $r_d\approx 2.1$, and the range of the strong force is set by the mass of the pion $m_\pi\approx 0.14\ \mathrm{GeV/c^2}$.
These naive cross sections are enhanced, however, by the existence of bound and unbound states. When such states are "shallow" (very close to threshold), their contribution to scattering is "universal" and does not depend on the details of the interaction generating the state. Their cross-section only depends on $\Delta$, the absolute value of the energy difference between threshold and the state:
$$
\sigma(state)\sim \frac{4 \pi \hbar^2}{2 \mu \Delta}
$$
where $\mu$ is the reduced mass of the initial two-body system. (See, for example, Weinberg's Lectures on Quantum Mechanics or the analysis of $pd$ scattering by Tumino et. al.)
Both the $np$ and $nnp$ systems each have one bound state, respectively the deuteron ($\Delta=B_d \approx 2.2\ \mathrm{MeV}$) and the triton ($\Delta=B_t \approx 6.3\ \mathrm{MeV}$), giving additional cross-sections
$$
\sigma_{np}\,(deuteron)\sim 2.3\ \mathrm{b}\\
\sigma_{nd}\,(triton) \sim 2.2\ \mathrm{b}
$$
The sum of the naive and bound state cross sections is consistent with the observed $nd$ cross-section of about $4$ b, but grossly underestimates the $np$ cross-section of about $20$ b.
This large cross-section is due to the $^1S_0$ $np$ virtual state. The nuclear force between two nucleons is stronger when their spins are parallel than anti-parallel, which is why the $^3S_1$ deuteron with parallel $n,p$ spins exists as a stable bound state, but the $^1S_0\ np$ state exists only as a virtual state with energy $\Delta_{np}\left(^1S_0\right) \sim 66$ keV above threshold.
$$
\sigma_{np}\,\mathrm{(\sim66\,\mathrm{keV}\,virtual)} \sim \frac{1}{4}\ \frac{4 \pi \hbar^2}{2 \mu_{np}\ \Delta_{np}\left(^1S_0\right) } \sim 20\ \mathrm{b}
$$
The factor of $1/4$ is because the final singlet state can only couple to one of the four possible ($J=0$ singlet, $J=1$ triplet) initial states.
Slow Neutron Capture
Because nucleons are so small and the range of the strong force is so short, with both length scales about a fermi, their interactions are primarily s-wave when neutron momenta are $p\lt \hbar/$fm, corresponding to neutron kinetic energies $E\lesssim 10$ MeV. Since the proton, neutron, deuteron, triton, and s-waves all have positive parity, electric dipole (E1) transitions (which have odd parity) are forbidden and even-parity magnetic dipole (M1) transitions dominate.
The initial and final states of any transition must be distinguishable and hence orthogonal. Since the deuteron is mostly ($\gtrsim95\%$) a triplet $^3S_1$ spin-1 state, the initial state in s-wave $np\rightarrow d \gamma$ capture must be a $^1S_0$ singlet. The spatial wave-functions of the initial $^1S_0$ and final $^3S_1$ states overlap almost perfectly, so a magnetic dipole $np\rightarrow d \gamma$ transition proceeds easily.
In contrast, as first noted by Schiff in 1937, similar symmetry considerations instead strongly inhibit $nd\rightarrow t \gamma$ transitions.
According to "Energy levels of light nuclei A=3", the $^3$H nuclei ground state wave function is mostly spatially symmetric S $(\sim 90\%)$ and D $(\sim 9\%)$ states, with a small $(\sim 1\%)$ mixed symmetry S' component, and $\lesssim 0.1\%$ P wave. (This article also lists many theoretical topics relevant to 3 nucleon systems.) There are negligible contributions from the completely antisymmetric $^2S_{1/2}$ state and from the 4 P states.
The initial spin-1/2 neutron and spin-1 deuteron can be in either a $^2S_{1/2}$ spin-1/2 doublet or $^4S_{3/2}$ spin-3/2 quartet state.
My rough (and possibly imperfect) understanding is that M1 transitions again can't couple initial and final $^2S$ states, so the only allowed M1 transitions are from the initial continuum $^4S$ state which is spatially anti-symmetric. Such transitions are only possible to the tiny mixed S' component of the triton, so the capture section is also tiny.
